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Let $X$ be a normal complex projective variety, let $V$ be a closed subset of $X$ (possibly reducible), and let $I_V$ be its ideal sheaf (consider the reduced scheme structure for example).

If $A$ is an ample divisor on $X$, then $\mathcal{O}_X(mA)\otimes I_V$ is globally generated if $m \gg 0$ by one of the definitions of ampleness. This implies in particular that there exists a divisor $A'\in |mA|$ such that $A'$ contains $V$ in its support. In fact there is much more than one divisor with this property.

In general there is no reason why $A'$ should be irreducible. In particular if $V$ contains at least two prime divisors, it is clear that every divisor containing $V$ will be reducible.

My question is the following:

Suppse $\dim V\leq \dim X-2$. Can I find, for $m \gg 0$, an irreducible divisor $A'\in |mA|$ containing $V$?

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2 Answers 2

up vote 7 down vote accepted

EDIT : Olivier Wittenberg pointed out to me that a positive answer to the question follows from Theorem 1 of [Altman-Kleiman, Bertini theorems for hypersurface sections containing a subscheme]. I keep below my previous answer (whose argument is different, but more complicated).


The answer to your question is positive. Let $X$ be an integral projective variety of dimension $n\geq 2$ (over an algebraically closed field) and $V\subset X$ a subvariety of codimension $\geq 2$. Let $A$ be ample on $X$ : up to replacing $A$ by a multiple, we may assume that $X\subset\mathbb{P}^N$ and $A=\mathcal{O}_X(1)$.

Let $\mathcal{H}_e$ be the projective space of degree $e$ hypersurfaces in $\mathbb{P}^N$, $\mathcal{F}_e^{igr}(X)\subset\mathcal{H}_e$ the subset consisting of hypersurfaces whose intersections with $X$ are not irreducible generically reduced of codimension $1$ in $X$, and $\mathcal{G}_e(V)\subset\mathcal{H}_e$ the subset consisting of hypersurfaces containing $V$.

The exact sequence $0\to\mathcal{I}_V(e)\to\mathcal{O}_{\mathbb{P}^N}(e)\to\mathcal{O}_V(e)\to 0$ shows that, when $ e\gg 0$, the codimension of $\mathcal{G}_e(V)$ in $\mathcal{H}_e$ is a polynomial of degree $\leq n-2$ in $e$ (the Hilbert polynomial of $V$). On the other hand, Théorème 0.4 of arXiv:0911.1118 shows that, when $e\geq 2$, the codimension of $\mathcal{F}_e^{igr}(X)$ in $\mathcal{H}_e$ is $\geq \binom{e+n-1}{n-1}-n$, that is at least a polynomial of degree $n-1$ in $e$. As a consequence, if $e \gg 0$, $\mathcal{G}_e(V)$ is not included in $\mathcal{F}_e^{igr}(X)$. A hypersurface in $\mathcal{G}_e(V)$ but not in $\mathcal{F}_e^{igr}(X)$ induces on $X$ the divisor you are looking for.

Note that if $X$ is moreover $S_2$ (for instance, normal), then this divisor is itself integral.

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Thanks Olivier! I really like this proof! –  Gianni Bello Jan 18 '13 at 9:54

EDIT 18/1 to make it clearer and somehow address the singular case:

First assume $X$ nonsingular. Denote $|mA-V|$ the linear system of divisors in $|mA|$ containing $V$. For $m\gg 0$, the base locus of $|mA-V|$ is exactly $V$ (blowing up the components of $V$ you can transform $|mA-V|$ into a base-point free system). So by Bertini (in characteristic zero), either you have an irreducible $A'$ or the image of $X$ by the corresponding map $f$ to projective space is ${\mathbb P}^1$. As diverietti says in the comments, the exact sequence in cohomology of $0 \rightarrow \mathcal{I}_V(mA) \rightarrow \mathcal{O}_X(mA) \rightarrow \mathcal{O}_V(mA)\rightarrow 0$ and Serre vanishing show that $\dim |mA-V|$ grows like $m^{\dim X}$. This being larger than 2 does not guarantee that the image of $f$ is not $\mathbb{P}^1$ (you could have $H^0(\mathcal{I}_V(mA))=H^0(f^*(\mathcal{O}_{\mathbb{P}^1}(k))$) but if that were the case, the rate of growth of $\dim |tmA-V|$ shows that for $t \gg 0$, $H^0(\mathcal{I}_V(tmA))$ strictly contains $H^0(f^*(\mathcal{O}_{\mathbb{P}^1}(tk))$ and so for some $t$ the system is not composed with a pencil.

In the singular case, let $\pi:Y \rightarrow X$ be a resolution of singularities such that $E=\pi^{-1}(\operatorname{Sing}(X))$ is a divisor, and assume that (*) for each component $V_i$ of $V$ there is an irreducible $\tilde V_i\subset Y$ of codimension at least 2 with $\pi(\tilde V_i)=V_i$. Let $\tilde A$ be an ample divisor of the form $\pi^*(kA)-D$ where $D$ is some divisor supported at $E$. The proof for the nonsingular case gives an irreducible divisor in $|m\tilde A-\tilde V|$ whose image in $X$ is irreducible and lies in $|kmA-V|$.

(*) is always true if no component of $V$ is contained in the singular locus, and I guess it is also true for components contained in the singular locus but don't know of a quick proof. On the other hand, Olivier's proof does not need characteristic zero, so unless someone asks, I won't try to polish this one further.

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I don't really understand. Which kind of Bertini are you applying? –  diverietti Jan 17 '13 at 12:07
    
Oh, sorry: I missed the "normal X" hypothesis, and assumed smooth. In that case, you can reduce this to Hartshorne's III,10.9 by blowing up (the components of) V. The proper transform of the linear system $|mA-V|$ is basepoint free. –  quim Jan 17 '13 at 13:25
    
Could you explain me a little bit better, please? I am sorry but I don't understand completely! Suppose $X$ is smooth. Coll $\mathfrak d_{m,V}$ the linear system of divisors in $|mA|$ which contain $V$. Bertini tells us that if $\mathfrak d_{m,V}$ has no base points then its generic element is non-singular (in fact irreducible and non singular provided the dimension of the image of the corresponding map to the projective space is at least two). So how do you conclude? And what about the reduction to the non singular case? –  diverietti Jan 17 '13 at 14:34
1  
Ok, I'll rewrite it more carefully asap (tonight possibly). About the reduction to the nonsingular case I am not 100% sure, I'll think of it. –  quim Jan 17 '13 at 15:18
    
Thanks quim! I'll wait for that! –  diverietti Jan 17 '13 at 15:37

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